Question: Solve for $n$, $ -\dfrac{3n - 1}{n^2} = -\dfrac{9}{4n^2} - \dfrac{1}{n^2} $
Explanation: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $n^2$ $4n^2$ and $n^2$ The common denominator is $4n^2$ To get $4n^2$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{3n - 1}{n^2} \times \dfrac{4}{4} = -\dfrac{12n - 4}{4n^2} $ The denominator of the second term is already $4n^2$ , so we don't need to change it. To get $4n^2$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ -\dfrac{1}{n^2} \times \dfrac{4}{4} = -\dfrac{4}{4n^2} $ This give us: $ -\dfrac{12n - 4}{4n^2} = -\dfrac{9}{4n^2} - \dfrac{4}{4n^2} $ If we multiply both sides of the equation by $4n^2$ , we get: $ -12n + 4 = -9 - 4$ $ -12n + 4 = -13$ $ -12n = -17 $ $ n = \dfrac{17}{12}$